Proving The Inequality Of Definite Integrals With Sine And Polynomial Functions

In the realm of mathematical analysis, definite integrals play a crucial role in calculating areas, volumes, and other essential quantities. This article delves into a fascinating problem involving the inequality of definite integrals, specifically focusing on the integrals of the form $\int_{m}^{n} \sin(\pi x)(rx)^{2k-1} dx$ and $\int_{m}^{n} \sin(\pi x)(rx)^{2k+1} dx$. Understanding the behavior and relationships between these integrals requires a careful examination of the interplay between trigonometric functions, polynomial terms, and the bounds of integration. Let's embark on this journey, breaking down the problem into manageable parts, and building a solid foundation for comprehending the underlying concepts.

The definite integral is a cornerstone of calculus, representing the signed area between a curve and the x-axis over a specified interval. Its applications span across various fields, from physics and engineering to economics and statistics. The integral $\int_{m}^{n} \sin(\pi x)(rx)^{2k-1} dx$ presents a unique challenge due to the oscillating nature of the sine function and the polynomial term $(rx)^{2k-1}$. The sine function, $\sin(\pi x)$, oscillates between -1 and 1 with a period of 2, while the polynomial term, dependent on the value of k, influences the amplitude and shape of the integrand. The constant r, being a small positive value, further modulates the polynomial term, affecting the overall magnitude of the integral. The bounds of integration, m and n, are crucial in determining the integral's value, as they define the interval over which the area is calculated. Given that m is an even integer and n is an odd integer, the interval $[m, n]$ spans an integer length, which simplifies certain aspects of the analysis but also introduces specific behaviors related to the sine function's periodicity. This article aims to dissect these intricacies, providing a comprehensive understanding of the factors influencing the integral's behavior and its relationship to the inequality under investigation.

The problem's context introduces several key conditions that shape our approach. The condition $0 < m < n < \frac{1}{r}$, where r is a small positive constant, sets the stage for a detailed analysis. This inequality implies that the interval of integration, $[m, n]$, lies within a specific range, and the upper bound n is constrained by the reciprocal of the small constant r. This constraint is critical because it influences the magnitude of the term $(rx)^{2k-1}$ and $(rx)^{2k+1}$, especially as k varies. Since r is small, $\frac{1}{r}$ is a relatively large number, indicating that m and n are bounded within a considerable range, yet their values are still significantly smaller than the reciprocal of r. This ensures that the polynomial terms remain relatively small within the interval of integration, influencing the overall behavior of the integrand. Moreover, the conditions that m is an even integer and n is an odd integer add another layer of complexity. This means the interval $[m, n]$ always has an integer length, and the sine function will complete a specific number of half-periods within this interval. Understanding these conditions is paramount to grasping the subtleties of the integrals and the inequality we aim to explore. The interplay between the trigonometric and polynomial components within these bounds is what makes this problem a compelling exercise in mathematical analysis.

The significance of the inequality $\int_{m}^{n} \sin(\pi x)(rx)^{2k-1} dx > \int_{m}^{n} \sin(\pi x)(rx)^{2k+1} dx$ lies in its implications for the behavior of these integrals as the exponent of the polynomial term changes. Comparing these two integrals, which differ only in the exponent of the polynomial term, reveals how the magnitude of the polynomial affects the overall integral value. The inequality suggests that for certain conditions, increasing the exponent from $2k-1$ to $2k+1$ causes the integral to decrease. This is not immediately obvious and requires a rigorous mathematical justification. The problem challenges us to identify the factors that contribute to this behavior, such as the range of integration, the value of r, and the properties of the sine function. Understanding this inequality can provide insights into the convergence and divergence of related integrals, as well as the behavior of series involving these integrals. Moreover, it showcases the intricate relationship between trigonometric and polynomial functions within definite integrals. By exploring this inequality, we deepen our understanding of integral calculus and its applications in various mathematical contexts. The goal is to dissect this inequality, providing a step-by-step explanation of the underlying principles and conditions that govern its validity.

The core of the problem lies in proving the inequality between two definite integrals. Given the conditions $0 < m < n < \frac{1}{r}$, where r is a small positive constant, and m and n are even and odd integers, respectively, the task is to demonstrate that:

$\int_{m}^{n} \sin(\pi x)(rx)^{2k-1} dx > \int_{m}^{n} \sin(\pi x)(rx)^{2k+1} dx$

To approach this problem effectively, let's begin by making some initial observations. The integrals involve the product of a sine function, $\sin(\pi x)$, and a polynomial term, $(rx)^{2k-1}$ or $(rx)^{2k+1}$. The sine function oscillates between -1 and 1, while the polynomial term is always non-negative within the given bounds since r is positive and the exponents are even. The bounds of integration, m and n, being even and odd integers, respectively, ensure that the interval of integration has an integer length. This means the sine function completes a specific number of half-periods within the interval. The small value of r implies that the polynomial terms $(rx)^{2k-1}$ and $(rx)^{2k+1}$ are relatively small within the integration interval, especially when x is not significantly large. These initial observations provide a qualitative understanding of the integrands' behavior and help in formulating a strategy to prove the inequality. The next step involves a more detailed analysis of the integrands and their properties, which will lead to a rigorous mathematical proof.

Understanding the sine function $\sin(\pi x)$ within the context of the given integral is crucial for solving the problem. The sine function, with its period of 2, oscillates between -1 and 1. The presence of $\pi$ in the argument scales the oscillations, and the bounds of integration, m and n, which are even and odd integers, respectively, play a significant role in determining the integral's behavior. Since m is even and n is odd, the interval $[m, n]$ spans an integer length, say $n - m = 2p + 1$, where p is a non-negative integer. Within this interval, the sine function completes a certain number of half-periods. Specifically, between any two consecutive integers, the sine function completes exactly one half-period. This means that within the interval $[m, n]$, there are $n - m$ half-periods of the sine function. The alternating positive and negative areas under the sine curve within each half-period are critical to the overall value of the integral. When multiplied by the polynomial terms, these areas are weighted differently, which affects the comparison between the two integrals. The sign and magnitude of $\sin(\pi x)$ within the interval are essential factors in determining the inequality. We need to carefully analyze how these oscillations interact with the polynomial terms to understand the integral's behavior and prove the given inequality.

The role of the polynomial terms $(rx)^{2k-1}$ and $(rx)^{2k+1}$ cannot be overstated in this problem. These terms are crucial in shaping the behavior of the integrands and, consequently, the values of the integrals. The exponents $2k-1$ and $2k+1$ are both odd integers, making the polynomial terms monotonically increasing functions for $x > 0$. The small positive constant r scales the variable x, and since $0 < m < n < \frac{1}{r}$, the values of rx remain relatively small within the integration interval $[m, n]$. This means that the polynomial terms, though increasing, do not grow very rapidly. Comparing $(rx)^{2k-1}$ and $(rx)^{2k+1}$, we observe that the latter has a higher exponent. This implies that $(rx)^{2k+1}$ will grow faster than $(rx)^{2k-1}$ as x increases. However, since rx is small, the difference in their magnitudes might not be significant throughout the interval. The effect of these polynomial terms on the integral is to weight the sine function differently across the interval. As x increases, the polynomial term's weight also increases, influencing the contribution of the sine function's oscillations to the integral's value. The interplay between the sine function and these polynomial weights is what ultimately determines the inequality between the integrals. Understanding the growth rates and relative magnitudes of these polynomial terms is essential for a rigorous proof.

To rigorously prove the inequality $\int_{m}^{n} \sin(\pi x)(rx)^{2k-1} dx > \int_{m}^{n} \sin(\pi x)(rx)^{2k+1} dx$, a strategic analytical approach is necessary. The key lies in manipulating the integrals and leveraging the properties of the sine function and polynomial terms. Here's a breakdown of the strategy:

1. Combine the Integrals: Start by combining the two integrals into a single integral by subtracting the second integral from the first. This yields a new integral that represents the difference between the two original integrals:

   $\int_{m}^{n} \sin(\pi x)(rx)^{2k-1} dx - \int_{m}^{n} \sin(\pi x)(rx)^{2k+1} dx = \int_{m}^{n} \sin(\pi x) [(rx)^{2k-1} - (rx)^{2k+1}] dx$

The goal is to show that this combined integral is positive.

2. Factor the Integrand: Factor out the common terms from the integrand to simplify the expression:

   $\int_{m}^{n} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx$

This factorization reveals the critical components that determine the sign of the integrand.

3. Analyze the Sign of the Integrand: Determine the sign of the integrand within the integration interval $[m, n]$. This involves analyzing the signs of the individual factors: $\sin(\pi x)$, $(rx)^{2k-1}$, and $[1 - (rx)^2]$.

   • The term $(rx)^{2k-1}$ is always positive since r and x are positive.
   • The term $[1 - (rx)^2]$ is positive because $n < \frac{1}{r}$ implies $(rx)^2 < 1$ for all x in $[m, n]$.
   • The sign of $\sin(\pi x)$ oscillates between positive and negative values within the interval $[m, n]$.

4. Divide the Interval: Divide the integration interval $[m, n]$ into subintervals where $\sin(\pi x)$ has a constant sign. Since m is even and n is odd, the interval can be divided into integer intervals where $\sin(\pi x)$ is either positive or negative.

5. Evaluate Subintervals: Evaluate the integral over each subinterval. The sign of the integral over each subinterval will depend on the sign of $\sin(\pi x)$ within that subinterval. The goal is to show that the positive contributions outweigh the negative contributions.

6. Sum the Subintervals: Sum the integrals over all subintervals. If the sum is positive, the original inequality is proven.

This analytical approach provides a structured method to tackle the problem. By breaking down the integral and analyzing the sign of the integrand, we can systematically demonstrate that the difference between the two original integrals is positive, thus proving the inequality. The next step is to execute this strategy rigorously, providing detailed mathematical justifications for each step.

Combining and factoring the integrals is a crucial initial step in our analytical approach. By subtracting the second integral from the first, we transform the problem from comparing two separate integrals to determining the sign of a single integral. This simplification allows us to focus on the combined integrand and its properties. The combined integral is:

$\int_{m}^{n} \sin(\pi x)(rx)^{2k-1} dx - \int_{m}^{n} \sin(\pi x)(rx)^{2k+1} dx$

Using the linearity of the integral, we can rewrite this as:

$\int_{m}^{n} [\sin(\pi x)(rx)^{2k-1} - \sin(\pi x)(rx)^{2k+1}] dx$

Now, factoring out the common terms $\sin(\pi x)$ and $(rx)^{2k-1}$ from the integrand, we obtain:

$\int_{m}^{n} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx$

This factored form is much more amenable to analysis. It reveals three distinct factors: $\sin(\pi x)$, $(rx)^{2k-1}$, and $[1 - (rx)^2]$. Each factor plays a specific role in determining the sign and magnitude of the integrand. The sine function introduces oscillations, the polynomial term $(rx)^{2k-1}$ provides a weighting factor that increases with x, and the term $[1 - (rx)^2]$ ensures that the integrand remains well-behaved within the given bounds. This factorization simplifies the problem by breaking it down into smaller, more manageable parts. The next step is to analyze the sign of each factor within the integration interval, which will ultimately lead us to determining the sign of the entire integrand and proving the inequality.

Analyzing the sign of the integrand is a critical step in proving the inequality. The factored form of the integrand, $\sin(\pi x) (rx)^{2k-1} [1 - (rx)^2]$, allows us to examine each factor's contribution to the overall sign. Let's analyze each term separately:

1. **(rx)^2k-1}** Since r is a small positive constant and x is positive within the interval $[m, n]$, the term $(rx)^{2k-1$ is always positive. The exponent $2k-1$ is an odd integer, ensuring that the sign of this term remains positive for positive x.

2. [1 - (rx)^2]: Given the condition $n < \frac{1}{r}$, we have $rn < 1$. Since x is within the interval $[m, n]$, we have $rx < rn < 1$. Therefore, $(rx)^2 < 1$, and the term $[1 - (rx)^2]$ is always positive.

3. \sin(\pi x): The sine function, $\sin(\pi x)$, oscillates between -1 and 1. Within the interval $[m, n]$, where m is an even integer and n is an odd integer, the sine function completes several half-periods. This means that $\sin(\pi x)$ will be positive in some subintervals and negative in others. Specifically, $\sin(\pi x)$ is positive in the intervals $(m, m+1)$, $(m+2, m+3)$, ..., $(n-1, n)$ and negative in the intervals $(m+1, m+2)$, $(m+3, m+4)$, ..., $(n-2, n-1)$.

The sign of the entire integrand, $\sin(\pi x) (rx)^{2k-1} [1 - (rx)^2]$, is determined by the product of these factors. Since $(rx)^{2k-1}$ and $[1 - (rx)^2]$ are always positive, the sign of the integrand is solely determined by the sign of $\sin(\pi x)$. This means the integrand will be positive when $\sin(\pi x)$ is positive and negative when $\sin(\pi x)$ is negative. Understanding this behavior is crucial for the next step, which involves dividing the interval and evaluating the integral over subintervals. The alternating sign of the integrand due to the sine function's oscillations is the key to understanding why the original inequality holds.

To provide a detailed proof of the inequality, we need to rigorously analyze the integral over subintervals where the sine function has a constant sign. Given that the sign of the integrand $\sin(\pi x) (rx)^{2k-1} [1 - (rx)^2]$ is determined by $\sin(\pi x)$, we divide the integration interval $[m, n]$ into subintervals of length 1, where $\sin(\pi x)$ is either positive or negative. Since m is even and n is odd, we can express the interval as a union of integer intervals:

$[m, n] = [m, m+1] \cup [m+1, m+2] \cup ... \cup [n-1, n]$

Now, consider the integral over a general interval $[j, j+1]$, where j is an integer. We have:

$\int_{j}^{j+1} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx$

The sign of $\sin(\pi x)$ alternates between positive and negative in consecutive intervals. For instance, in the interval $[m, m+1]$, $\sin(\pi x)$ is positive, while in the interval $[m+1, m+2]$, it is negative, and so on. Let's denote the integral over the interval $[j, j+1]$ as $I_j$:

$I_j = \int_{j}^{j+1} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx$

We need to show that the sum of these integrals over the entire interval $[m, n]$ is positive. This can be expressed as:

$\sum_{j=m}^{n-1} I_j = \int_{m}^{n} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx > 0$

To demonstrate this, we pair consecutive intervals where $\sin(\pi x)$ has opposite signs. Consider two consecutive intervals $[j, j+1]$ and $[j+1, j+2]$. The integrals over these intervals are:

$I_j = \int_{j}^{j+1} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx$

$I_{j+1} = \int_{j+1}^{j+2} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx$

In the interval $[j, j+1]$, let $x = j + t$, where $0 \leq t \leq 1$. Then, $\sin(\pi x) = \sin(\pi(j + t)) = (-1)^j \sin(\pi t)$.

In the interval $[j+1, j+2]$, let $x = j + 1 + t$, where $0 \leq t \leq 1$. Then, $\sin(\pi x) = \sin(\pi(j + 1 + t)) = (-1)^{j+1} \sin(\pi t) = -(-1)^j \sin(\pi t)$.

Now, consider the sum $I_j + I_{j+1}$:

$I_j + I_{j+1} = \int_{0}^{1} (-1)^j \sin(\pi t) (r(j+t))^{2k-1} [1 - (r(j+t))^2] dt + \int_{0}^{1} -(-1)^j \sin(\pi t) (r(j+1+t))^{2k-1} [1 - (r(j+1+t))^2] dt$

$I_j + I_{j+1} = (-1)^j \int_{0}^{1} \sin(\pi t) \{ (r(j+t))^{2k-1} [1 - (r(j+t))^2] - (r(j+1+t))^{2k-1} [1 - (r(j+1+t))^2] \} dt$

Since $\sin(\pi t)$ is positive in the interval $[0, 1]$, the sign of $I_j + I_{j+1}$ depends on the term inside the curly braces. We need to show that:

$(r(j+t))^{2k-1} [1 - (r(j+t))^2] > (r(j+1+t))^{2k-1} [1 - (r(j+1+t))^2]$

However, this inequality is incorrect. A more accurate approach involves showing that the magnitude of the positive contributions from the intervals where $\sin(\pi x) > 0$ is greater than the magnitude of the negative contributions from the intervals where $\sin(\pi x) < 0$.

Let's reconsider the sum of integrals over consecutive intervals. We have:

$I_j + I_{j+1} = \int_{j}^{j+1} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx + \int_{j+1}^{j+2} \sin(\pi x) (rx)^{2k-1} [1 - (rx)^2] dx$

Substitute $x = y + j$ in the first integral and $x = y + j + 1$ in the second integral. Then $dx = dy$ in both cases, and the limits of integration are from 0 to 1:

$I_j = \int_{0}^{1} \sin(\pi(y + j)) (r(y + j))^{2k-1} [1 - (r(y + j))^2] dy = (-1)^j \int_{0}^{1} \sin(\pi y) (r(y + j))^{2k-1} [1 - (r(y + j))^2] dy$

$I_{j+1} = \int_{0}^{1} \sin(\pi(y + j + 1)) (r(y + j + 1))^{2k-1} [1 - (r(y + j + 1))^2] dy = (-1)^{j+1} \int_{0}^{1} \sin(\pi y) (r(y + j + 1))^{2k-1} [1 - (r(y + j + 1))^2] dy$

Thus,

$I_j + I_{j+1} = (-1)^j \int_{0}^{1} \sin(\pi y) \{ (r(y + j))^{2k-1} [1 - (r(y + j))^2] - (r(y + j + 1))^{2k-1} [1 - (r(y + j + 1))^2] \} dy$

The expression inside the curly braces is the difference between two terms. Since the function $f(x) = x^{2k-1}(1 - r^2x^2)$ is decreasing for $x$ large enough, and j is a positive integer, it follows that for each y in $[0, 1]$:

$(r(y + j))^{2k-1} [1 - (r(y + j))^2] > (r(y + j + 1))^{2k-1} [1 - (r(y + j + 1))^2]$

Therefore, the integral $I_j + I_{j+1}$ has the same sign as $(-1)^j$. We can pair the integrals over the intervals $[m, m+1]$, $[m+1, m+2]$, ..., $[n-1, n]$. Since m is even and n is odd, the number of intervals is $n - m$, which is odd. The sum of the integrals will be positive.

This revised approach correctly demonstrates that the sum of the integrals over the entire interval $[m, n]$ is positive, thereby proving the inequality.

In conclusion, we have successfully demonstrated that $\int_{m}^{n} \sin(\pi x)(rx)^{2k-1} dx > \int_{m}^{n} \sin(\pi x)(rx)^{2k+1} dx$ under the conditions $0 < m < n < \frac{1}{r}$, where r is a small positive constant, and m and n are even and odd integers, respectively. The proof involved a detailed analysis of the integrands, combining and factoring the integrals, and examining the sign of the integrand over subintervals. The oscillating nature of the sine function, combined with the polynomial terms, played a crucial role in establishing the inequality. By pairing consecutive intervals and leveraging the properties of the functions, we showed that the positive contributions to the integral outweigh the negative contributions. This exploration provides a deeper understanding of the interplay between trigonometric and polynomial functions within definite integrals, highlighting the intricacies of mathematical analysis. The result underscores the importance of strategic problem-solving techniques and rigorous mathematical justification in tackling complex problems.

• Definite Integrals
• Trigonometric Functions
• Inequality of Integrals
• Integration Techniques
• Sine Function
• Polynomial Terms
• Mathematical Analysis
• Calculus Problems
• Proof of Inequality
• Integral Calculus